Imagine A is a NxM matrix with two basic properties
1) Each element in the matrix is distinct and lies in the range of 1<=A[i][j]<=(N*M)
2) For any two cells of the matrix, (i1,j1) and (i2,j2), if (i1^j1) > (i2^j2) then A[i1][j1] > A[i2][j2] ,where
1 ≤ i1,i2 ≤ N
1 ≤ j1,j2 ≤ M.
^ is Bitwise XOR
Given N and M , you have to calculatethe total number of matrices of size N x M which have both the properties
mentioned above.
Input format:
First line contains T, the number of test cases. 2*T lines follow with N on the first line and M on the second, representing the number of rows and columns respectively.
Output format:
Output the total number of such matrices of size N x M. Since, this answer can be large, output it modulo 10^9+7
Constraints:
1 ≤ N,M,T ≤ 1000
SAMPLE INPUT
1
2
2
SAMPLE OUTPUT
4
Explanation
The four possible matrices are:
[1 3] | [2 3] | [1 4] | [2 4]
[4 2] | [4 1] | [3 2] | [3 1]
题意：找出有多少个排列矩阵 使得下标i1^j1>j2^j2 对应的元素也为a[i1][j1]>a[i2][j2]
排列矩阵是包含1-n*m每个数的矩阵
题解：先考虑一个矩阵n*m的答案，我们只要把i^j相同的位置做排列就行了。因为他们两个的大小没法比，所以可以排列。因为i^j<=1024 所以记录数字的多少开1024即可。然后考虑t个n*m，我们可以处理出行的前缀和
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<deque> using namespace std; typedef long long ll; ll f[1200],sum[1005][1050],vis[1050]; int main() { ll i,j,t,n,m; f[0]=1; for(i=1; i<=1024; i++)//计算排列的种类，例如总共有3个地方i^j相同的，总共有几种排列方法，那就是上一个的排列方法*3 { f[i]=f[i-1]*i00000007; } for(i=1; i<=1000; i++)//计算到最后一行的时候，每一种相同的i^j的个数 { for(j=1; j<=1000; j++) { sum[i][i^j]++; } for(j=0; j<=1024; j++) sum[i][j]+=sum[i-1][j]; } scanf("%lld",&t); while(t--) { scanf("%lld%lld",&n,&m); memset(vis,0,sizeof(vis)); ll ans=1; for(i=0; i<=1024; i++)//到第n行时，把每一种i^j的数量赋值给vis vis[i]=sum[n][i]; for(i=1; i<=n; i++)//因为我们vis【i】存的是n=n，m=1000，每种i^j的总个数，但是我们求得时候是用的n,m《=1000，所以只需要把总的渐去后面的就是当前的 { for(j=m+1; j<=1000; j++) { vis[i^j]--; } } for(i=0; i<=1024; i++)//求总的情况 ans=ans*f[vis[i]]00000007; printf("%lld\n",ans); } return 0; }
